Resolve into Partial Fractions
(iv) `(x+3)/((1-x)^(2)(1+x^(2)))`

Answer

To resolve the expression \(\frac{x+3}{(1-x)^2(1+x^2)}\) into partial fractions, we will follow these steps:

### Step 1: Set up the Partial Fraction Decomposition
We start by expressing the given fraction in terms of its partial fractions. Since the denominator consists of a repeated linear factor \((1-x)^2\) and an irreducible quadratic factor \((1+x^2)\), we can write:

\[ \frac{x+3}{(1-x)^2(1+x^2)} = \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{Cx + D}{1+x^2} \]

### Step 2: Multiply through by the Denominator
Next, we multiply both sides of the equation by the denominator \((1-x)^2(1+x^2)\) to eliminate the fractions:

\[ x + 3 = A(1-x)(1+x^2) + B(1+x^2) + (Cx + D)(1-x)^2 \]

### Step 3: Expand the Right Side
Now we expand the right side:

1. For \(A(1-x)(1+x^2)\):
\[ A(1-x)(1+x^2) = A(1 + x^2 - x - x^3) = A(1 - x + x^2 - x^3) \]

2. For \(B(1+x^2)\):
\[ B(1+x^2) = B + Bx^2 \]

3. For \((Cx + D)(1-x)^2\):
\[ (Cx + D)(1 - 2x + x^2) = Cx(1 - 2x + x^2) + D(1 - 2x + x^2) = Cx - 2Cx^2 + Cx^3 + D - 2Dx + Dx^2 \]

Combining all these, we get:

\[ x + 3 = (A + B + D) + (C - 2D - 2C)x + (A + B - 2C + D)x^2 + (-A + C)x^3 \]

### Step 4: Collect Like Terms
Now we collect like terms based on the powers of \(x\):

- Coefficient of \(x^3\): \(-A + C\)
- Coefficient of \(x^2\): \(A + B - 2C + D\)
- Coefficient of \(x\): \(-2D + C\)
- Constant term: \(A + B + D\)

### Step 5: Set Up the System of Equations
Now we can equate the coefficients from both sides:

1. For \(x^3\): \(-A + C = 0\) (1)
2. For \(x^2\): \(A + B - 2C + D = 0\) (2)
3. For \(x\): \(-2D + C = 1\) (3)
4. For the constant term: \(A + B + D = 3\) (4)

### Step 6: Solve the System of Equations
From equation (1), we have \(C = A\).

Substituting \(C = A\) into equations (2), (3), and (4):

From (2):
\[ A + B - 2A + D = 0 \implies -A + B + D = 0 \implies B + D = A \quad (5) \]

From (3):
\[ -2D + A = 1 \implies A - 2D = 1 \quad (6) \]

From (4):
\[ A + B + D = 3 \quad (7) \]

Now we have a system of three equations (5), (6), and (7).

### Step 7: Substitute and Solve
From (5), we can express \(B\) in terms of \(A\) and \(D\):
\[ B = A - D \quad (8) \]

Substituting (8) into (7):
\[ A + (A - D) + D = 3 \implies 2A = 3 \implies A = \frac{3}{2} \]

Now substituting \(A = \frac{3}{2}\) into (6):
\[ \frac{3}{2} - 2D = 1 \implies -2D = 1 - \frac{3}{2} \implies -2D = -\frac{1}{2} \implies D = \frac{1}{4} \]

Now substituting \(A\) and \(D\) back into (5):
\[ B + \frac{1}{4} = \frac{3}{2} \implies B = \frac{3}{2} - \frac{1}{4} = \frac{6}{4} - \frac{1}{4} = \frac{5}{4} \]

And since \(C = A\):
\[ C = \frac{3}{2} \]

### Step 8: Write the Partial Fraction Decomposition
Now we have:
- \(A = \frac{3}{2}\)
- \(B = \frac{5}{4}\)
- \(C = \frac{3}{2}\)
- \(D = \frac{1}{4}\)

Thus, the partial fraction decomposition is:

\[ \frac{3/2}{1-x} + \frac{5/4}{(1-x)^2} + \frac{(3/2)x + (1/4)}{1+x^2} \]

### Final Answer
The resolved partial fractions are:

\[ \frac{3/2}{1-x} + \frac{5/4}{(1-x)^2} + \frac{(3/2)x + (1/4)}{1+x^2} \]